将由逗号分隔的数字组成的用户输入放入数组中？

Question

I am taking in 10 numbers from the user (user enters them at a prompt, and the numbers are separated by commas, as so: 245645, -243, 4245). How can I put these elements into an array? As shown below, I have used scanf which does not work as I had hoped. Any suggestions would be appreciated.

//User will pass ten numbers, separated by commas. This is to be put into an array.

#include <stdio.h>

int main () 
{
    int A[10]; // array to contain in users input.
    printf("Enter your numbers: ");
    scanf("%d", &A[10]);

    return 0;
}

Answer 1

You have to consume the comma as well in scanf :

for(int i=0; i<10; i++) {        /* for each element in A */
  if(0==scanf("%d,", &A[i])) {    /* read a number from stdin into A[i] and then consume a commma (if any) */
    break;                       /* if no digit was read, user entered less than 10 numbers separated by commas */
    /* alternatively error handling if fewer than 10 is illegal */
  }
}

Answer 2

I won't write the whole thing for you. But I can definitely help. One of the ways to do that will be:

1. Get a string that contains 10 comma-separated numbers: fgets() may be?
2. Validate the string, trim white-spaces as well, makes life easier
3. Pick out a number from string: strtol() may be?
4. Search for a ',' character in the string, and set pointer to the next index after ',' : strchr() may be?
5. Repeat steps 3 and 4 for a total of 10 times (from here, 9 times actually)
6. Print the numbers

The code below would do half of your job. The only remaining part would be to get string from user and validate it. The intention to have a string declared and initialised upfront is to put more emphasise on actual parsing of data which appear complicated to beginners (no offence).

Before we look at the code below, lets read a few things first.

• You might want to take a look at the man page for strtol() function
• You might want to take a look at the man page for fgets() function, which is not used in the code below, but you may end-up using it to achieve step 1.

I already concede the fact that this may not be the best way to achieve it, and I would happily agree that this code below can be made better in thousand ways by adding various error check, but I leave that to you to explore and implement.

#include<stdio.h>
#include<stdlib.h>
#include<string.h>

int main(void) 
{
    int i = 0, j = 0;
    char *Str = "1,11,21,1211,111221,312211,1234,4321,123,789";
    char *ptr;
    long ret;
    long array[10] = { [0 ... 9] = 0 };

    // So, lets assume step 1 and 2 are taken care of.

    // Iterate though the data for 10 times
    for(i = 0; i < 10; i++)
    {
        ret = strtol(Str, &ptr, 10);
        // Check if its a number
        if(ret != 0)
        {
            // Its is a number!
            // Put it in the array
            array[j] = ret;
            // Increment the index so that next number will not over-write the existing one
            j++;
            // Find the next ',' in the string  
            Str = strchr(Str, ',');
            if(Str == NULL)
            {
                // Handle this condition that there are no more commas in the string! 
                break;
            }
            // Assuming that the ',' is found, increment the pointer to index next to ','
            Str++;
        }
    }

    // Print the array
    for(i = 0; i < j; i++)
        printf("%ld\n", array[i]);
}

This prints the following output:

1
11
21
1211
111221
312211
1234
4321
123
789

Hope I have got you started, Good luck.